In this chapter, we provide RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs for English medium students, Which will very helpful for every student in their exams. Students can download the latest RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs Maths pdf, free RD Sharma Solutions for Chapter 12 Heron’s Formula MCQs Maths book pdf download. Now you will get step by step solution to each question.

Textbook | NCERT |

Class | Class 9 |

Subject | Maths |

Chapter | Chapter 12 |

Chapter Name | Heron’s Formula |

Exercise | MCQs |

**RD Sharma Solutions for Class 9 Chapter 12 Heron’s Formula MCQs Download PDF**

Question 1.

In ∆ABC ≅ ∆LKM, then side of ∆LKM equal to side AC of ∆ABC is

(a) LX

(b) KM

(c) LM

(d) None of these

Solution:

Side AC of ∆ABC = LM of ∆LKM (c)

Question 2.

In ∆ABC ≅ ∆ACB, then ∆ABC is isosceles with

(a) AB=AC

(b) AB = BC

(c) AC = BC

(d) None of these

Solution:

∵ ∆ABC ≅ ∆ACB

∴ AB = AC (a)

Question 3.

In ∆ABC ≅ ∆PQR, then ∆ABC is congruent to ∆RPQ, then which of the following is not true:

(a) BC = PQ

(b) AC = PR

(c) AB = PQ

(d) QR = BC

Solution:

∵ ∆ABC = ∆PQR

∴ AB = PQ, BC = QR and AC = PR

∴ BC = PQ is not true (a)

Question 4.

In triangles ABC and PQR three equality relations between some parts are as follows: AB = QP, ∠B = ∠P and BC = PR State which of the congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS

Solution:

In two triangles ∆ABC and ∆PQR,

AB = QP, ∠B = ∠P and BC = PR

The condition apply : SAS (a)

Question 5.

In triangles ABC and PQR, if ∠A = ∠R, ∠B = ∠P and AB = RP, then which one of the following congruence conditions applies:

(a) SAS

(b) ASA

(c) SSS

(d) RHS

Solution:

In ∆ABC and ∆PQR,

∠A = ∠R

∠B = ∠P

AB = RP

∴ ∆ABC ≅ ∆PQR (ASA axiom) (b)

Question 6.

If ∆PQR ≅ ∆EFD, then ED =

(a) PQ

(b) QR

(c) PR

(d) None of these

Solution:

∵ ∆PQR = ∆EFD

∴ ED = PR (c)

Question 7.

If ∆PQR ≅ ∆EFD, then ∠E =

(a) ∠P

(b) ∠Q

(c) ∠R

(d) None of these

Solution:

∵ ∆PQR ≅ ∆EFD

∴ ∠E = ∠P (a)

Question 8.

In a ∆ABC, if AB = AC and BC is produced to D such that ∠ACD = 100°, then ∠A =

(a) 20°

(b) 40°

(c) 60°

(d) 80°

Solution:

In ∆ABC, AB = AC

∴ ∠B = ∠C

But Ext. ∠ACD = ∠A + ∠B

∠ACB + ∠ACD = 180° (Linear pair)

∴ ∠ACB + 100° = 180°

⇒ ∠ACB = 180°-100° = 80°

∴ ∠B = ∠ACD = 80°

But ∠A + ∠B 4- ∠C = 180°

∴ ∠A + 80° + 80° = 180°

⇒∠A+ 160°= 180°

∴ ∠A= 180°- 160° = 20° (a)

Question 9.

In an isosceles triangle, if the vertex angle is twice the sum of the base angles, then the measure of vertex angle of the triangle is

(a) 100°

(b) 120°

(c) 110°

(d) 130°

Solution:

In ∆ABC,

∠A = 2(∠B + ∠C)

= 2∠B + 2∠C

Adding 2∠A to both sides,

∠A + 2∠A = 2∠A + 2∠B + 2∠C

⇒ 3∠A = 2(∠A + ∠B + ∠C)

⇒ 3∠A = 2 x 180° (∵∠A + ∠B + ∠C = 180° )

⇒ 3∠A = 360°

⇒∠A = 360∘3 = 120°

∴ ∠A = 120° (b)

Question 10.

Which of the following is not a criterion for congruence of triangles?

(a) SAS

(b) SSA

(c) ASA

(d) SSS

Solution:

SSA is not the criterion of congruence of triangles. (b)

Question 11.

In the figure, the measure of ∠B’A’C’ is

(a) 50°

(b) 60°

(c) 70°

(d) 80°

Solution:

In the figure,

∆ABC ≅ ∆A’B’C’

∴ ∠A = ∠A

⇒3x = 2x- + 20

⇒ 3x – 2x = 20

⇒ x = 20

∠B’A’C’ = 2x + 20 = 2 x 20 + 20

= 40 + 20 = 60° (b)

Question 12.

If ABC and DEF are two triangles such that ∆ABC ≅ ∆FDE and AB = 5 cm, ∠B = 40° and ∠A = 80°. Then, which of the following is true?

(a) DF = 5 cm, ∠F = 60°

(b) DE = 5 cm, ∠E = 60°

(c) DF = 5 cm, ∠E = 60°

(d) DE = 5 cm, ∠D = 40°

Solution:

∵ ∆ABC ≅ ∆FDE,

AB = 5 cm, ∠A = 80°, ∠B = 40°

∴ DF = 5 cm, ∠F = 80°, ∠D = 40°

∴ ∠C =180°- (80° + 40°) = 180° – 120° = 60°

∴ ∠E = ∠C = 60°

∴ DF = 5 cm, ∠E = 60° (c)

Question 13.

In the figure, AB ⊥ BE and FE ⊥ BE. If BC = DE and AB = EF, then ∆ABD is congruent to

(a) ∆EFC

(b) ∆ECF

(c) ∆CEF

(d) ∆FEC

Solution:

In the figure, AB ⊥ BE, FE ⊥ BE

BC = DE, AB = EF,

then CD + BC = CD + DE BD = CE

In ∆ABD and ∆CEF,

BD = CE (Prove)

AB = FE (Given)

∠B = ∠E (Each 90°)

∴ ∆ABD ≅ ∆FCE (b)

Question 14.

In the figure, if AE || DC and AB = AC, the value of ∠ABD is

(a) 70°

(b) 110°

(c) 120°

(d) 130°

Solution:

In the figure, AE || DC

∴ ∠1 = 70° (Vertically opposite angles)

∴ ∠1 = ∠2 (Alternate angles)

∠2 = ∠ABC (Base angles of isosceles triangle)

∴ ABC = 90°

But ∠ABC + ∠ABD = 180° (Linear pair)

⇒ 70° +∠ABD = 180°

⇒∠ABD = 180°-70°= 110°

∴ ∠ABD =110° (b)

Question 15.

In the figure, ABC is an isosceles triangle whose side AC is produced to E. Through C, CD is drawn parallel to BA. The value of x is

(a) 52°

(b) 76°

(c) 156°

(d) 104°

Solution:

In ∆ABC, AB = AC

AC is produced to E

CD || BA is drawn

∠ABC = 52°

∴ ∠ACB = 52° (∵ AB = AC)

∴ ∠BAC = 180°-(52° +52°)

= 180°-104° = 76°

∵ AB || CD

∴ ∠ACD = ∠BAC (Alternate angles)

= 76°

and ∠BCE + ∠DCB = 180° (Linear pair)

∠BCE + 52° = 180°

⇒∠BCE = 180°-52°= 128°

∠x + ∠ACD = 380°

⇒ x + 76° = 180°

∴ x= 180°-76°= 104° (d)

Question 16.

In the figure, if AC is bisector of ∠BAD such that AB = 3 cm and AC = 5 cm, then CD =

(a) 2 cm

(b) 3 cm

(c) 4 cm

(d) 5 cm

Solution:

In the figure, AC is the bisector of ∠BAD, AB = 3 cm, AC = 5 cm

In ∆ABC and ∆ADC,

AC = AC (Common)

∠B = ∠D (Each 90°)

∠BAC = ∠DAC (∵ AC is the bisector of ∠A)

∴ ∆ABC ≅ ∆ADC (AAS axiom)

∴ BC = CD and AB = AD (c.p.c.t.)

Now in right ∆ABC,

AC^{2} = AB^{2} + BC^{2}

⇒ (5)^{2} = (3)^{2} + BC^{2}

⇒25 = 9 + BC^{2}

⇒ BC^{2} = 25 – 9 = 16 = (4)^{2}

∴ BC = 4 cm

But CD = BC

∴ CD = 4 cm (c)

Question 17.

D, E, F are the mid-point of the sides BC, CA and AB respectively of ∆ABC. Then ∆DEF is congruent to triangle

(a) ABC

(b) AEF

(c) BFD, CDE

(d) AFE, BFD, CDE

Solution:

In ∆ABC, D, E, F are the mid-points of the sides BC, CA, AB respectively

DE, EF and FD are joined

∵ E and F are the mid-points

AC and AB,

∴ EF = 12 BC and EF || BC

Similarly,

DE = 12 AB and DE || AB

DF = 12 AC and DF || AC

∴ ∆DEF is congruent to each of the triangles so formed

∴ ∆DEF is congruent to triangle AFE, BFD, CDE (d)

Question 18.

ABC is an isosceles triangle such that AB = AC and AD is the median to base BC. Then, ∠BAD =

(a) 55°

(b) 70°

(c) 35°

(d) 110°

Solution:

In ∆ABC, AB = AC

AD is median to BC

∴ BD = DC

In ∆ADB, ∠D = 90°, ∠B = 35°

But ∠B + BAD + ∠D = 180° (Sum of angles of a triangle)

⇒ 35° + ∠BAD + 90° = 180°

⇒∠BAD + 125°= 180°

⇒ ∠BAD = 180°- 125°

⇒∠BAD = 55° (a)

Question 19.

In the figure, X is a point in the interior of square ABCD. AXYZ is also a square. If DY = 3 cm and AZ = 2 cm, then BY =

(a) 5 cm

(b) 6 cm

(c) 7 cm

(d) 8 cm

Solution:

In the figure, ABCD and AXYZ are squares

DY = 3 cm, AZ = 2 cm

DZ = DY + YZ

= DY + Z = 3 + 2 = 5 cm

In ∆ADZ, ∠2 = 90°

AD^{2} + AZ^{2} + DZ^{2} = 2^{2} + 5^{2} cm

= 4 + 25 = 29

In ∠ABX, ∠X = 90°

AB^{2} = AX^{2} + BX^{2}

AD^{2} = AZ^{2} + BX^{2}

(∵ AB = AD, AX = AZ sides of square)

29 = 2^{2} + BX^{2}

⇒ 29 = 4 + BX^{2}

⇒ BX^{2} = 29 – 4 = 25 = (5)^{2}

∴ BX = 5 cm (a)

Question 20.

In the figure, ABC is a triangle in which ∠B = 2∠C. D is a point on side BC such that AD bisects ∠BAC and AB = CD. BE is the bisector of ∠B. The measure of ∠BAC is

(a) 72°

(b) 73°

(c) 74°

(d) 95°

Solution:

In the figure, ∠B = 2∠C, AD and BE are the bisectors of ∠A and ∠B respectively,

AB = CD

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